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**Objective Type Questions****Q.35. The sides of an equilateral triangle are increasing at the rate of 2 cm/sec. The rate at which the area increases, when side is 10 cm is:****(a) 10 cm ^{2}/s **

Let the length of each side of the given equilateral triangle be x cm.

âˆ´

Area of equilateral triangle

âˆ´

Hence, the rate of increasing of area = 10 âˆš3 cm

Hence, the correct option is (c).

(d) 10 radian/sec

Length of ladder = 5 m

Let AB = y m and BC = x m

âˆ´ In right Î”ABC,

AB

â‡’ x^{2} + y^{2} = (5)^{2} â‡’ x^{2} + y^{2} = 25

Differentiating both sides w.r.t x, we have

â‡’

â‡’[âˆµ x = 2m]

â‡’

â‡’x (- 0.1) = 0

â‡’

Now(Î¸ is in radian)

â‡’

Differentiating both sides w.r.t. t, we get

â‡’

radian/sec

[(â€“) sign shows the decrease of change of angle]

Hence, the required rate =radian/sec

Hence, the correct option is (b).**Q.37. The curve y =x ^{1/5} has at (0, 0) **

Equation of curve is y = x

Differentiating w.r.t. x, we get

(at x = 0)

âˆ´ The tangent is parallel to y-axis.

Hence, the correct option is (a).

Given equation of the curve is 3x

Differentiating both sides w.r.t. x, we get

is the slope of the tangent

âˆ´ Slope of the normal =

Now x + 3y = 8 is parallel to the normal

Differentiating both sides w.r.t. x, we have

âˆ´

Putting y = x in eq. (i) we get

3x

âˆ´ x = Â± 2 and y = Â± 2

So the points are (2, 2) and (â€“ 2, â€“ 2).

Equation of normal to the given curve at (2, 2) is

â‡’ 3y â€“ 6 = â€“ x + 2 â‡’ x + 3y â€“ 8 = 0

Equation of normal at (â€“ 2, â€“ 2) is

â‡’ 3y + 6 = â€“ x â€“ 2 â‡’ x + 3y + 8 = 0

âˆ´ The equations of the normals to the curve are

x + 3y Â± 8 = 0

Hence, the correct option is (c).

Q.39. If the curve ay + x

Equation of the given curves are ay + x

and x

Differentiating eq. (i) w.r.t. x, we have

âˆ´

Now differentiating eq. (ii) w.r.t. x, we get

The two curves are said to be orthogonal if the angle between the tangents at the point of intersection is 90Â°.

(1, 1) is the point of intersection of two curves.

âˆ´ 6(1)

So a = 6

Hence, the correct option is (d).

Solution.

Given that y = x

Î”x = 2.00 â€“ 1.99 = 0.01

âˆ´

= 4 x (2)^{3} x 0.01 = 32 x 0.01 = 0.32

Hence, the correct option is (a).**Q.41. The equation of tangent to the curve y (1 + x ^{2}) = 2 â€“ x, where it crosses x-axis is:**

Solution.

Given that y(1 + x

If it cuts x-axis, then y-coordinate is 0.

âˆ´ 0(1 + x

Put x = 2 in equation (i)

y(1 + 4) = 2 â€“ 2 â‡’ y(5) = 0 â‡’ y = 0

Point of contact = (2, 0)

Differentiating eq. (i) w.r.t. x, we have

Equation of tangent is y â€“ 0 =

â‡’ 5y = â€“ x + 2 â‡’ x + 5y = 2

Hence, the correct option is (a).

Solution.

Given that y = x

Differentiating both sides w.r.t. x, we have

â‡’

Since the tangents are parallel to x-axis, then

âˆ´ 3x

âˆ´ y

y

âˆ´ Points are (2, 2) and (â€“ 2, 34)

Hence, the correct option is (d).

Solution.

Equation of the curve is y = e

Slope of the tangent

âˆ´ Equation of tangent to the curve at (0, 1) is

y â€“1 = 2(x â€“ 0)

â‡’ y â€“ 1 = 2x â‡’ y â€“ 2x = 1

Since the tangent meets x-axis where y = 0

âˆ´ 0 â€“ 2x = 1 â‡’

So the point is

Hence, the correct option is (b).

Q.44. The slope of tangent to the curve x = t

Solution.

The given curve is x = t

âˆ´

Now (2, â€“ 1) lies on the curve

â‡’ 2 = t

â‡’ t

â‡’ t(t + 5) â€“ 2(t + 5) = 0

â‡’ (t + 5) (t â€“ 2) = 0

âˆ´ t = 2, t = â€“ 5 and â€“ 1 = 2t

â‡’ 2t

â‡’ t

â‡’ t(t â€“ 2) + 1 (t â€“ 2) = 0 â‡’ (t + 1) (t - 2) = 0

â‡’ t = â€“ 1 and t = 2

So t = 2 is common value

âˆ´

Hence, the correct option is (b).

Solution.

The given curves are x

and 3x

Differentiating eq. (i) w.r.t. x, we get

â‡’

âˆ´

So slope of the curve

Differentiating eq. (ii) w.r.t. x, we get

âˆ´

So the slope of the curve

Now

So the angle between the curves is

Hence, the correct option is (c).

Solution.

The given function is f (x) = 2x

f â€²(x) = 6x

For increasing and decreasing f â€²(x) = 0

âˆ´ 6x

â‡’ x

â‡’ x(x + 2) + 1(x + 2) = 0 â‡’ (x + 2) (x + 1) = 0

â‡’ x = â€“ 2, x = â€“ 1

The possible intervals are (â€“âˆž, - 2), ( - 2, - 1), ( - 1, âˆž)

Now f â€²(x) = (x + 2) (x + 1)

â‡’ f â€²(x)( - âˆž, - 2) = (-) (-) = (+) increasing

â‡’ f â€²(x)(- 2, - 1) = (+) (-) = (-) decreasing

â‡’ f â€²(x)(- 1, âˆž) = (+) (+) = (+) increasing

Hence, the correct option is (b).

Solution.

Given that f (x) = 2x + cos x

f â€²(x) = 2 - sin x

Since f â€²(x) > 0 âˆ€ x

So f (x) is a n increasing function.

Hence, the correct option is (d).

Solution.

Here y = x(x â€“ 3)

For increasing and decreasing

âˆ´ 2x(x â€“ 3) + (x â€“ 3)2 = 0 â‡’ (x - 3) (2x + x - 3) = 0

â‡’ (x â€“ 3) (3x â€“ 3) = 0 â‡’ 3(x - 3) (x - 1) = 0

âˆ´ x = 1, 3

âˆ´ Possible intervals are (â€“ âˆž, 1), (1, 3), (3, âˆž)

dy/dx = (x - 3) (x - 1)

For (- âˆž , 1) =(-) (-) = (+) increasing

For (1, 3) =(-) (+) = (-) decreasing

For (3, âˆž) =(+) (+) = (+) increasing

So the function decreases in (1, 3) or 1 < x < 3

Hence, the correct option is (a).

Solution.

Here,

f (x) = 4 sin

f â€²(x) = 12 sin

= 12 cos x [sin

= 12 cos x [sin

âˆ´ 1 - sin x â‰¥ 0 and sin

âˆ´ sin

Hence, f â€²(x) > 0 , when cos x > 0 i.e.,

So, f(x) is in creasing whereand f â€²(x) < 0

when cos x < 0 i.e.

Hence, (x) is decreasing when

So f(x) is decreasing in

Hence, the correct option is (b).

Solution.

Here, Let f (x) = cos x; So, f â€²(x) = - sin x

f â€²(x) < 0 in

So f (x) = cos x is decreasing in

Hence, the correct option is (c).

Solution.

Here, f(x) = tan x â€“ x So, f â€²(x) = sec

f â€²(x) > 0 âˆ€ x âˆˆ R

So f (x) is always increasing

Hence, the correct option is (a).

Solution.

Let f (x) = x

f â€²(x) = 2x - 8

For local maxima and local minima, f â€²(x) = 0

âˆ´ 2x â€“ 8 = 0 â‡’ x = 4

So, x = 4 is the point of local maxima and local minima.

f â€³(x) = 2 > 0 minima at x = 4

âˆ´ f(x)

= 16 â€“ 32 + 17 = 33 â€“ 32 = 1

So the minimum value of the function is 1

Hence, the correct option is (c).

Solution.

Let f(x) = x

For local maxima and local minima f â€²(x) = 0

âˆ´ 3x

â‡’ x

â‡’ x(x - 8) - 4(x - 8) = 0 â‡’ (x - 8) (x - 4) = 0

âˆ´ x = 8, 4 âˆˆ [0, 9]

So, x = 4, 8 are the points of local maxima and local minima.

Now we will calculate the absolute maxima or absolute

minima at x = 0, 4, 8, 9

âˆ´ f(x) = x

f(x)

f(x)

= 64 â€“ 288 + 384 = 448 â€“ 288 = 160

f(x)

= 512 â€“ 1152 + 768 = 1280 â€“ 1152 = 128

f(x)

= 729 â€“ 1458 + 864 = 1593 â€“ 1458 = 135

So, the absolute minimum value of f is 0 at x = 0

Hence, the correct option is (b).

Solution.

We have f(x) = 2x

f â€²(x) = 6x

For local maxima and local minima f â€²(x) = 0

âˆ´ 6x

â‡’ x

â‡’ x(x â€“ 2) + 1(x â€“ 2) = 0 â‡’ (x + 1) (x - 2) = 0

â‡’ x = -1, 2 are the points of local maxima and local minima

Now f â€³(x) = 12x - 6

f â€³(x)

f â€³(x)

So, the function is maximum at x = -1 and minimum at x = 2

Hence, the correct option is (c).

(d) 2âˆš2

Solution.

We have f (x) = sin x cos x

â‡’

â‡’ f â€²(x) = cos 2x

Now for local maxima and local minima f â€²(x) = 0

âˆ´ cos 2x = 0

â‡’

âˆ´

f â€³(x) = - 2 sin 2x

So f (x) is maximum at

âˆ´ Maximum value of f(x) =

Hence, the correct option is (b).

Solution.

We have f (x) = 2 sin 3x + 3 cos 3x

f â€²(x) = 2 cos 3x Ã— 3 - 3 sin 3xÃ—3 = 6 cos 3x - 9 sin 3x

f â€³(x) = - 6 sin 3x Ã— 3 - 9 cos 3x Ã— 3

= - 18 sin 3x - 27 cos 3x

=â€“ 18 < 0

maxima Maximum value of f(x) at

Hence, the correct option is (a).

Solution.

Given that y = â€“ x

âˆ´ Slope of the given curve,

m = â€“ 3x

For local maxima and local minima,

âˆ´â€“ 6x + 6 = 0 â‡’ x = 1

Now

Maximum value of the slope at x = 1 is

mx = 1 = â€“ 3(1)2 + 6(1) + 9 = â€“ 3 + 6 + 9 = 12

Hence, the correct option is (b).

(b)x = 1/e

(d) x =âˆše

Solution.

We have f (x) = x

Taking log of both sides, we have

log f (x) = x log x

Differentiating both sides w.r.t. x, we get

â‡’ f â€²(x) = f (x) [1 + log x] = x

To find stationary point, f â€²(x) = 0

âˆ´ x

x

â‡’ log x = â€“ 1 x = e

Hence, the correct option is (b).

Solution.

Let

Taking log on both sides, we get

â‡’ log [f (x)] = x log x

Differentiating both sides w.r.t. x, we get

â‡’

For local maxima and local minima f â€²(x) = 0

âˆ´ 1 + log x = 0 â‡’ log x = â€“ 1 â‡’ x = e

So,is the stationary point.

Now

âˆ´ Maximum value of the function at

Hence, the correct option is (c).

Fill in the blanks

We have

y = 4x

and y = x

Differentiating eq. (i) w.r.t. x, we have

[m is the slope of curve (i)]

Differentiating eq. (ii) w.r.t. x, we get

= 3x

[m

If the two curves touch each other, then m

âˆ´ 8x + 2 = 3x

â‡’ 3x

â‡’ 3x(x - 3) + 1(x - 3) = 0 â‡’ (x - 3) (3x + 1) = 0

âˆ´

Putting x = 3 in eq. (i), we get

y = 4(3)

So, the required point is (3, 34)

Now for x =

âˆ´ Other required point is

Hence, the required points are (3, 34) and

We have y = tan x. So,

âˆ´ Slope of the normal =

at the point (0, 0) the slope = â€“ cos

So the equation of normal at (0, 0) is y â€“ 0 = â€“1(x â€“ 0)

â‡’ y = â€“ x â‡’ y + x =0

Hence, the required equation is y + x = 0.

We have f (x) = sin x â€“ ax + b â‡’ f â€²(x) = cos x - a

For increasing the function f â€²(x) > 0

âˆ´ cos x - a > 0

Since cos x âˆˆ [ - 1, 1]

âˆ´ a < â€“1 â‡’ a âˆˆ ( - âˆž, - 1)

Hence, the value of a is ( - âˆž, - 1).

We have f (x) =

For decreasing the function f â€²(x) < 0

âˆ´

âˆ´ x > Â± 1 â‡’ x âˆˆ (1, âˆž)

Hence, the required interval is (1, âˆž).

Here,

For maximum and minimum value f â€²(x) = 0

âˆ´

Now

(âˆµ a, b > 0)

Hence,

minima So the least value of the function at x

Hence, least value is

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